Bolding Cone Head

And if we add Circular Standing Wave Resonance to the minimum $KE$ from the post "Two Quantum Wells, Quantum Tunneling, $v_{min}$" dated !9 Jul 2015.

This standing wave resonance occurs when the frequency of oscillation $f_{SHM}$ (SHM along a radial line), is an integer multiple of the angular frequency, $f_{cir}$.  The results of resonance are sharp peaks at discrete locations along $x$.

When we flip this graph vertically,

which is the absorption spectrum below a threshold intensity.

$f_{cir,\,1}=\cfrac{v_{min,\,1}}{2\pi x_1}$

$f_{cir,\,2}=\cfrac{v_{min,\,2}}{2\pi x_2}$

But,

$\cfrac{f_{SHM}}{f_{cir,\,1}}=N$

$\cfrac{f_{SHM}}{f_{cir,\,2}}=M$

which leads to the proportion identity,

$\cfrac{x_1}{x_2}\cfrac{v_{min,\,2}}{v_{min,\,1}}=\cfrac{N}{M}$

For three dip points on the graph, $X_1,\,\,Y_1$, $X_2,\,\,Y_2$ and $X_3,\,\,Y_3$,

$\cfrac{X_1}{X_2}\sqrt{\cfrac{Y_2}{Y_1}}=\cfrac{N}{M}$

$\cfrac{X_1}{X_3}\sqrt{\cfrac{Y_3}{Y_1}}=\cfrac{N}{O}$

$\cfrac{X_3}{X_2}\sqrt{\cfrac{Y_2}{Y_3}}=\cfrac{O}{M}$

From which $f_{SHM}$ can be calculated, once $M$, $N$ or $O$ has been determined graphically.

You are not going to get a graph of $v^2_{min}$ versus $x$, of course.

But in my dream...